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Musya8 [376]
3 years ago
6

What is the modulus of resilience for a tensile test specimen with a nearly linear elastic region if the yield strength is 500MP

a at a strain of 0.003
Engineering
2 answers:
Temka [501]3 years ago
4 0

Answer:

U = 0.75 MPa

Therefore, the modulus of resilience for the tensile test specimen is 0.75MPa

Explanation:

The Modulus of resilience can be derived from the area under the elastic portion of the stress-strain diagram. It can be written mathematically as;

U = 1/2 × yield strength × yield strain.

Where U is the modulus of resilience.

Given;

Yield strength = 500MPa

Yield strain = 0.003

Substituting the given values we have;

U = 1/2 × 500MPa × 0.003

U = 0.75 MPa

Therefore, the modulus of resilience for the tensile test specimen is 0.75MPa

ella [17]3 years ago
3 0

Answer:

The modulus of resilience is 166.67 MPa

Explanation:

Modulus of resilience is given by yield strength ÷ strain

Yield strength = 500 MPa

Strain = 0.003

Modulus of resilience = 500 MPa ÷ 0.003 = 166.67 MPa

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Explanation:

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<u>At state 1:</u>

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The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

<u>At state 2:</u>

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 \mathbf{m^{3}} / kg

From temperature T₂ = 200⁰ C

v_f_2 = 0.0016 \ m^3/kg  

vg_2 = 0.127 \ m^3/kg  

Since  vf_2 < v_2 , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}

x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}

x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}

\mathsf{x_2 =0.78}

At temperature T₂, the specific internal energy u_f_2 = 850.6 \ kJ/kg , also ug_2 = 2594.3 \ kJ/kg

Thus,

u_2 = uf_2 + x_2 (ug_2 -uf_2)

u_2 =850.6  +0.78 (2594.3 -850.6)

u_2 =850.6  +1360.086

u_2 =2210.686 \ kJ/kg

<u>At state 3:</u>

Temperature T_3=T_2 = 200 ^0 C ,

V_3 = 2V_1 = 0.06 \ m^3

Specific volume v_3 = 0.2  \ m^3/kg

Thus; vg_3 =vg_2 = 0.127 \ m^3/kg ,

SInce v_3 > vg_3, therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at v_3 = 0.2  \ m^3/kg and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 \ m^3/kg

The specific internal energy u_3 at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

u_2-u_1

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

u_3-u_2

= (2622.3 - 2210.686)  kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg  

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

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